Circle And System Of Circles Question 259
Question: Any circle through the point of intersection of the lines $ x+\sqrt{3}y=1 $ and $ \sqrt{3}x-y=2 $ if intersects these lines at points P and Q, then the angle subtended by the arc PQ at its centre is
[MP PET 1998]
Options:
A) $ 180^{o} $
B) $ 90^{o} $
C) $ 120^{o} $
D) Depends on centre and radius
Show Answer
Answer:
Correct Answer: A
Solution:
Let the point of intersection of two lines is A.
$ \therefore $ The angle subtended by PQ on centre C $ = $ Two times the angle subtended by PQ on point A.
For $ x+\sqrt{3}y=1 $ , $ m_1=\frac{-1}{\sqrt{3}} $ and For $ \sqrt{3}x-y=2, $
$ m_2=\sqrt{3} $
$ \because $ $ m_1\times m_2=\frac{-1}{\sqrt{3}}\times \sqrt{3}=-1 $ ,
$ \therefore \ \angle A=90^{o} $
$ \therefore $ The angle subtended by arc PQ at its centre $ =2\times 90^{o}=180^{o} $
Trick: Given lines are perpendicular to each other, so PQ passes through centre of circle,
Hence arc makes $ 180^{o} $ to centre.
BETA
